Find the equation of the tangent to the curve y = (5x+4)/(3x -8) at the point (2, -7)

(Much easier to explain in conversation and with drawings): But, 

An application of the quotient rule, stating that for f(x)=a(x)/b(x), f'(x) = a'(x)b(x) - a(x)b'(x) all over [b(x))]^2 Here, a(x) = 5x + 4 and b(x) = 3x - 8

Thus applying the rule we find that: f'(x) = [5(3x-8) - 3(5x + 4)] / (3x - 8)^2 As we have been given a point on this line, we can substitue in x = 2

This gives f'(x) = (-10 - 42) / 4 f'(x) = -52/4 f'(x) = -13

Given that we have the pint 2, -7 we can substitute these values, and our found gradient to work out the equation of the line at the tangent with the model of y = mx + c

-7 = -13(2) + c

c = 19

Thus the equation of the line at the tangent is y = -13x + 19